Question

(x+1)(x+9)(x+5)^2+63

Answer 1

Step  1  :

1 Simplify —— x2

Equation at the end of step  1  :

1 (x + 9) • ((—— - 2x) - 63) = 0 x2

Step  2  :

Rewriting the whole as an Equivalent Fraction :

 2.1   Subtracting a whole from a fraction 

Rewrite the whole as a fraction using  x2  as the denominator :

2x 2x • x2 2x = —— = ——————— 1 x2

Equivalent fraction : The fraction thus generated looks different but has the same value as the whole 

Common denominator : The equivalent fraction and the other fraction involved in the calculation share the same denominator

Adding fractions that have a common denominator :

 2.2       Adding up the two equivalent fractions 
Add the two equivalent fractions which now have a common denominator

Combine the numerators together, put the sum or difference over the common denominator then reduce to lowest terms if possible:

1 - (2x • x2) 1 - 2x3 ————————————— = ——————— x2 x2

Equation at the end of step  2  :

(1 - 2x3) (x + 9) • (————————— - 63) = 0 x2

Step  3  :

Rewriting the whole as an Equivalent Fraction :

 3.1   Subtracting a whole from a fraction 

Rewrite the whole as a fraction using  x2  as the denominator :

63 63 • x2 63 = —— = ——————— 1 x2

Trying to factor as a Difference of Cubes:

 3.2      Factoring:  1 - 2x3 

Theory : A difference of two perfect cubes,  a3 - b3 can be factored into
              (a-b) • (a2 +ab +b2)

Proof :  (a-b)•(a2+ab+b2) =
            a3+a2b+ab2-ba2-b2a-b3 =
            a3+(a2b-ba2)+(ab2-b2a)-b3 =
            a3+0+0+b3 =
            a3+b3

Check :  1  is the cube of  1 

Check :  2  is not a cube !! 
Ruling : Binomial can not be factored as the difference of two perfect cubes



Adding fractions that have a common denominator :

 3.4       Adding up the two equivalent fractions 

(1-2x3) - (63 • x2) -2x3 - 63x2 + 1 ——————————————————— = ——————————————— x2 x2

Equation at the end of step  3  :

(-2x3 - 63x2 + 1) (x + 9) • ————————————————— = 0 x2

Step  4  :

Step  5  :

Pulling out like terms :

 5.1     Pull out like factors :

   -2x3 - 63x2 + 1  =   -1 • (2x3 + 63x2 - 1) 

Polynomial Roots Calculator :

 5.2    Find roots (zeroes) of :       F(x) = 2x3 + 63x2 - 1

     See theory in step 3.3 
In this case, the Leading Coefficient is  2  and the Trailing Constant is  -1. 

 The factor(s) are: 

of the Leading Coefficient :  1,2 
 of the Trailing Constant :  1 

 Let us test ....

  P  Q  P/Q  F(P/Q)   Divisor     -1     1      -1.00      60.00        -1     2      -0.50      14.50        1     1      1.00      64.00        1     2      0.50      15.00   


Polynomial Roots Calculator found no rational roots

Step  6  :

Pulling out like terms :

 6.1     Pull out like factors :

   -x - 9  =   -1 • (x + 9) 

Equation at the end of step  6  :

(-x - 9) • (2x3 + 63x2 - 1) ——————————————————————————— = 0 x2

Step  7  :

When a fraction equals zero :

 7.1    When a fraction equals zero ...

Where a fraction equals zero, its numerator, the part which is above the fraction line, must equal zero.

Now,to get rid of the denominator, Tiger multiplys both sides of the equation by the denominator.

Here's how:

(-x-9)•(2x3+63x2-1) ——————————————————— • x2 = 0 • x2 x2

Now, on the left hand side, the  x2  cancels out the denominator, while, on the right hand side, zero times anything is still zero.

The equation now takes the shape :
   (-x-9)  •  (2x3+63x2-1)  = 0

Theory - Roots of a product :

 7.2    A product of several terms equals zero. 

 When a product of two or more terms equals zero, then at least one of the terms must be zero. 

 We shall now solve each term = 0 separately 

 In other words, we are going to solve as many equations as there are terms in the product 

 Any solution of term = 0 solves product = 0 as well.

Solving a Single Variable Equation :

 7.3      Solve  :    -x-9 = 0 

 Add  9  to both sides of the equation : 
                      -x = 9 
Multiply both sides of the equation by (-1) :  x = -9 


Cubic Equations :

 7.4     Solve   2x3+63x2-1 = 0

Future releases of Tiger-Algebra will solve equations of the third degree directly.

Meanwhile we will use the Bisection method to approximate one real solution.

).

The function is   F(x) = 2x3 + 63x2 - 1

At   x=   -32.00   F(x)  is equal to  -1025.00 
At   x=   -31.00   F(x)  is equal to  960.00 

Intuitively we feel, and justly so, that since  F(x) is negative on one side of the interval, andpositive on the other side then, somewhere inside this interval,  F(x)  is zero 

Procedure :
(1) Find a point "Left" where F(Left) < 0

(2) Find a point 'Right' where F(Right) > 0

(3) Compute 'Middle' the middle point of the interval [Left,Right]

(4) Calculate Value = F(Middle)

(5) If Value is close enough to zero goto Step (7)

Else :  
If Value < 0 then : Left <- Middle
If Value > 0 then : Right <- Middle

(6) Loop back to Step (3)

(7) Done!! The approximation found is Middle




     Next Middle will get us close enough to zero: 

     F( -0.126241334 ) is  -0.000000700  

     The desired approximation of the solution is: 

       x ≓ -0.126241334 

     Note, ≓ is the approximation symbol 

Two solutions were found :

       x ≓ -0.126241334 x = -9

Answer 2

Answer:

Step-by-step explanation:

I think this question need 3 hours to write.